How do you test for convergence of #Sigma 5/(6n^2+n-1)# from #n=[1,oo)#?

Question

Evelyn Agard · Accepted Answer

#sum_(n=1)^oo 5/(6n^2+n-1)# is convergent.

As the terms of the series are positive we can use the integral test, using: #f(x) = 5/(6x^2+x-1)# We have that: #f(x) > 0# for #x in [1,+oo)]# #f'(x) = - frac (5(12x+1)) ((6x^2+x-1)^2) < 0 # for #x in [1,+oo]#, so the function is monotone decreasing in that interval. #lim_(x->oo) 5/(6x^2+x-1) = 0# #f(n) = 5/(6n^2+n-1)# so all the hypotheses of the integral test are satisfied and we can calculate: #int_1^oo (5dx)/(6x^2+x-1)# using partial fractions: #6x^2 +x -1 = (2x+1)(3x-1)# #A/(2x+1) + B/(3x-1) = 5/((2x+1)(3x-1))# #A(3x-1)+B(2x+1) = 5# #(3A+2B)x^2 -(A-B) = 5# #{ color(white) [ color(black) ((3A+2B = 0) ,( A-B = -5) ) color(white) ] color(black)# #{ color(white) [ color(black) ((A=-2 ),( B=3) ) color(white) ] color(black)# So: #int_1^oo (5dx)/(6x^2+x-1) = int_1^oo (-2dx)/(2x+1) +int_1^oo (3dx)/(3x-1) = [ln(3x-1) - ln (2x+1)]_1^oo# Using the properties of logarithms: #ln(3x-1) - ln(2x+1) = ln ((3x-1)/(2x+1))# so that: #[ln(3x-1) - ln (2x+1)]_1^oo = [ln ((3x-1)/(2x+1))]_1^oo# And as: # lim_(x->oo) ln ((3x-1)/(2x+1)) = ln(3/2)# #[ln ((3x-1)/(2x+1))]_1^oo = ln(3/2) - ln (2/3) = 2ln(3/2)# Finally we have: #int_1^oo (5dx)/(6x^2+x-1) = 2ln(3/2)# and as the integral is convergent, so is the series.

Ezra Adams · Answer

undefined To test for the convergence of the series $ \sum \frac{5}{6n^2 + n - 1} $ from $ n = 1 $ to infinity:

1. Use the Limit Comparison Test or Direct Comparison Test with a known convergent or divergent series to determine convergence.
2. Alternatively, use the Ratio Test or Root Test to examine the convergence of the series.